08年中考数学圆解答题选

学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网2008中考数学圆解答题选1.（08辽宁沈阳）21．如图所示，AB是O的一条弦，ODAB，垂足为C，交O于点D，点E在O上．（1）若52AOD，求DEB的度数；（2）若3OC，5OA，求AB的长．解：（1）ODAB，ADDB……3分11522622DEBAOD……5分（2）ODAB，ACBC，AOC△为直角三角形，3OC，5OA，由勾股定理可得2222534ACOAOC·········································8分28ABAC····················································································10分2.（08辽宁十二市）20.如图10，AB为O的直径，D为弦BE的中点，连接OD并延长交O于点F，与过B点的切线相交于点C．若点E为AF的中点，连接AE．求证：ABEOCB△≌△．解：（1）证明：如图2．AB是O的直径．90E··························································1分又BC是O的切线，90OBCEOBC·····················································3分OD过圆心，BDDE，EFFBBOCA．·····················································································6分E为AF中点，EFBFAE30ABE·························································································8分EBDCAO第21题图图10ODBCFEA学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网90E12AEABOB··················································································9分ABEOCB△≌△．·············································································10分3.(08北京市卷19题)19．（本小题满分5分）已知：如图，在RtABC△中，90C，点O在AB上，以O为圆心，OA长为半径的圆与ACAB，分别交于点DE，，且CBDA．（1）判断直线BD与O的位置关系，并证明你的结论；（2）若:8:5ADAO，2BC，求BD的长．解：（1）直线BD与O相切．……1分证明：如图1，连结OD．OAOD，AADO．90C，90CBDCDB．又CBDA，90ADOCDB．90ODB．直线BD与O相切．············································································2分（2）解法一：如图1，连结DE．AE是O的直径，90ADE．:8:5ADAO，4cos5ADAAE．·················································································3分90C，CBDA，4cos5BCCBDBD．·········································································4分2BC，52BD．·································································5分DCOABEDCOABE图1学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网解法二：如图2，过点O作OHAD于点H．12AHDHAD．:8:5ADAO，4cos5AHAAO．··············3分90C，CBDA，4cos5BCCBDBD．···························4分2BC，52BD．····························································································5分4.（08天津市卷）21．（本小题8分）如图，在梯形ABCD中，AB∥CD，⊙O为内切圆，E为切点，（Ⅰ）求AOD的度数；（Ⅱ）若8AOcm，6DOcm，求OE的长．解（Ⅰ）∵AB∥CD，∴180ADCBAD．……1分∵⊙O内切于梯形ABCD，∴AO平分BAD，有BADDAO21，DO平分ADC，有ADCADO21．∴90)(21ADCBADADODAO．∴90)(180ADODAOAOD．·····················································4分（Ⅱ）∵在Rt△AOD中，8AOcm，6DOcm，∴由勾股定理，得1022DOAOADcm．·············································5分∵E为切点，∴ADOE．有90AEO．··················································6分∴AODAEO．又OAD为公共角，∴△AEO∽△AOD．················································7分∴ADAOODOE，∴8.4ADODAOOEcm．······················································8分DCOABH图2ABDCEO学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网CABEFMN图①CABEFMN图②5.（08天津市卷）25．（本小题10分）已知Rt△ABC中，90ACB，CBCA，有一个圆心角为45，半径的长等于CA的扇形CEF绕点C旋转，且直线CE，CF分别与直线AB交于点M，N．（Ⅰ）当扇形CEF绕点C在ACB的内部旋转时，如图①，求证：222BNAMMN；思路点拨：考虑222BNAMMN符合勾股定理的形式，需转化为在直角三角形中解决．可将△ACM沿直线CE对折，得△DCM，连DN，只需证BNDN，90MDN就可以了．请你完成证明过程：（Ⅱ）当扇形CEF绕点C旋转至图②的位置时，关系式222BNAMMN是否仍然成立？若成立，请证明；若不成立，请说明理由．25．本小题满分10分．（Ⅰ）证明将△ACM沿直线CE对折，得△DCM，连DN，则△DCM≌△ACM．········································································1分有CACD，AMDM，ACMDCM，ACDM．又由CBCA，得CBCD．······························2分由DCMDCMECFDCN45，ACMECFACBBCNACMACM454590，得BCNDCN．················································································3分又CNCN，∴△CDN≌△CBN．··········································································4分有BNDN，BCDN．∴90BACDNCDMMDN．···············································5分CABEFDMN学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网∴在Rt△MDN中，由勾股定理，得222DNDMMN．即222BNAMMN．···········································6分（Ⅱ）关系式222BNAMMN仍然成立．···············································7分证明将△ACM沿直线CE对折，得△GCM，连GN，则△GCM≌△ACM．········································8分有CACG，AMGM，ACMGCM，CAMCGM．又由CBCA，得CBCG．由45GCMECFGCMGCN，ACMACMECFACNACBBCN45)(90．得BCNGCN．·············································································9分又CNCN，∴△CGN≌△CBN．有BNGN，45BCGN，135180CABCAMCGM，∴9045135CGNCGMMGN．∴在Rt△MGN中，由勾股定理，得222GNGMMN．即222BNAMMN．···········································10分6.（08内蒙赤峰）24．（本题满分14分）如图（1），两半径为r的等圆1O和2O相交于MN，两点，且2O过点1O．过M点作直线AB垂直于MN，分别交1O和2O于AB，两点，连结NANB，．（1）猜想点2O与1O有什么位置关系，并给出证明；（2）猜想NAB△的形状，并给出证明；（3）如图（2），若过M的点所在的直线AB不垂直于MN，且点AB，在点M的两侧，那么（2）中的结论是否成立，若成立请给出证明．CABEFMNG学科网(ZXXK.COM)-学海泛舟系列资料上学科网，下精品资料！学科网-学海泛舟系列资料版权所有@学科网24．解：（1）2O在1O上····························（1分）证明：2O过点1O，12OOr．又1O的半径也是r，点2O在1O上．···································（3分）（2）NAB△是等边三角形·························（5分）证明：MNAB，90NMBNMA．BN是2O的直径，AN是1O的直径，即2BNANr，2O在BN上，1O在AN上．······································（7分）连结12OO，则12OO是NAB△的中位线．1222ABOOr．ABBNAN，则NAB△是等边三角形．········································（9分）（3）仍然成立．·······